What is \; {dx}{²(^{-1} x)} equal to? [NDA-I 2021]

Correct Answer is : tan⁡−1x+c^{-1} x + ctan−1x+c

Correct Answer is : tan⁡−1x+c^{-1} x + ctan−1x+c

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UPSC NDA I 2021 Mathematics Solved Paper

Question : 101 of 120

Marks: +1, -0

What is

\int\; \frac{dx}{\sec^{2}(\tan^{-1} x)}

Solution: 👈: Video Solution

CONCEPT:

\sec^{2} x - \tan^{2} x = 1

\int\; \frac{dx}{1+x^{2}} = \tan^{-1} x + C \;\;

where

C

is a constant

CALCULATION:

Let

I = \int\; \frac{dx}{\sec^{2}(\tan^{-1} x)}

As we know that,

\sec^{2} x - \tan^{2} x = 1

\Rightarrow I = \int\; \frac{dx}{1+\tan^{2}(\tan^{-1} x)}

\Rightarrow I = \int\; \frac{dx}{1+[\tan(\tan^{-1} x)]^{2}}

\Rightarrow I = \int\; \frac{dx}{1+x^{2}}

As we know that,

\int\; \frac{dx}{1+x^{2}} = \tan^{-1} x + C

where

C

is constant.

\Rightarrow I = \int\; \frac{dx}{\sec^{2}(\tan^{-1} x)} = \tan^{-1} x + C

Hence, option 2 is correct.

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